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Unfortunately i reckon that the papers are getting more and more easy. Mind you not by much though.

I don't believe that taeyang used max height as the point where the projectile clears the wall. The max height is just used to help define variables of the parabola to then determine V and /alpha using the information given as the projectile clears the wall.

I wouldnt say mechanics is hard per se, it just requires a very good knowledge of the physics concepts, there's absolutely no difficult maths. Definitely harder 3u is the worst, have a look at most Q8, they most likely have some stupid double inequality....hate those!

You are such a white kid, and here is why:

I think they found x = 27 (which i found as well) you then need to sub in and find the maximum y value *tested and confirmed: maximum y value (y = 15.125) at x = 27

For anyone: 1. Copy Movie file onto a USB drive 2. Put USB in PS3 3. Watch movie thanks to PS3 divx codec :D If you even more lazy: Use windows media player to set up a media stream, which will show up on the PS3

Im at a loss as to why people print out past papers...? Okay you can take them around with you, but i find no problem in sitting at a desk with a laptop with a digital copy. o.O + its eco friendly.

Just going to skip to the second step: 2. Let 7^k(1+3k)-1=9M Where M is an integer Prove true for n = k+1 7^{k+1}(1+3(k+1))-1 7.7^k(4+3k)-1 7(7^k(1+3k)-1+1+3.7^k)-1 Using result from 2. 7(9M+1+3.7^k)-1 But since 7^k(1+3k)-1=9M 7^k = \frac{9M+1}{1+3k} Now subbing back in...

Yeah but capacity just sounds stupid.

Yeah, sure why not....there is no way im going to get that probability Q.

:( circle geo and now probability, how do you know my weaknesses......i vote hax!

\angle cab + \angle abc + \angle bca = 180^o (1) (interior angle sum of triangle) \angle egd + \angle egf + \angle fgd = 360^o (2) (Angle sum at a point) However since EGDC is a cyclic quad \angle egd = 180^o -\angle bca (3) Similiarly for BDGF \angle fgd = 180^o -\angle abc (4) Subbing 3 and...

Just sub in : x + 1/x and solve the polynomial.... EDIT: nvm that doesnt work....ive resorted to finding a + 1/a + b + 1/b + c + 1/c ...etc etc btw if your doing the further questions: 6....the answer is 2x^3 + 3x^2 + 8 = 0 (from the book)

its actually any mass that travels at the speed of light would have an infinite mass. as v approaches c, m_o is arbitrary whilst m_v becomes infinite m_v =\frac{m_o}{0}

But your not accounting for the part of the cylinder that extrudes the sphere...

Re: Simple Harmonic Motion --> Trouble understanding velocity expressions -- help nee I think an OHSNAP! is in order.

http://www.mediafire.com/?urmz36xv0z5t9km creds go to ohexploitable for uploading :D still looking for solutions tho...

Does anyone have the solutions for the 2011 SBHS MX1 trial??

Med Tsunami :S?

Just wondering....if there are so many students applying for medicine and such, how come Australia has a shortage of doctors?